Reactive Transport Mechanism for Organic Oxidation during

Dec 6, 2011 - School of Engineering and Applied Sciences, Harvard University, ... Similar to batch electrolysis, mass transfer, physical adsorption, a...
0 downloads 0 Views 180KB Size
Reactive Transport Mechanism for Organic Oxidation during Electrochemical Filtration: Mass-Transfer, Physical Adsorption, and Electron-Transfer Supporting Information Journal of Physical Chemistry Revised November 28, 2011 Han Liu and Chad D. Vecitis* *

Corresponding author: Chad D. Vecitis, Email: [email protected], Phone: (617) 496-1458. Address: School of Engineering and Applied Sciences, Harvard University, Cambridge, MA 02138

Figure S1. Electrochemical filtration apparatus. A) Design of the modified commercial polycarbonate filtration casing consisting of 1) a perforated stainless steel cathode, 2) an insulating silicone rubber electrode separator and seal, 3) a titanium anodic ring that is pressed into the carbon nanotube anodic filter, and 4) the MWNT anodic filter supported by a PTFE membrane. B, C) Images of the modified filtration casing. D, E) Images of the MWNT network before and after electrochemical filtration, respectively.

1 (-)

Influent

+

2 1



SS 3 Cathode

e-

e-

1 2 3 4

B

Effluent A

3 (+)

D

C

4

4

Prepared Anodic Filter

Used Anodic Filter

E

Figure S2. Effect of temperature on dye adsorption to the CNTs. Adsorption used 0.015 g CNTs, V = 100 mL, and allowed for 24 h to reach equilibrium. The points are experimental data and lines are fits to the Langmuir isotherm.

-1

Sorption amount (mg g )

40 35 30 25 20 15 o

15 C o 25 C o 35 C

10 5 0 0

50

100

150

200

Equilibrium concentration (µM)

250

Table S1. Langmuir Isotherm Parameters for MO Adsorption onto CNTs. Langmuir constants A

T

Thermodynamic parameters B

b

qm

(L mg-1)

(mg g-1)

15

2.81

32.2

0.977

-35.6

25

2.33

28.5

0.983

-36.3

35

2.07

25.3

0.996

-37.2

(oC)

A

R2

qe = qmCe/(1/b+Ce) ;

B

∆G°

∆H°

∆S°

(kJ mol-1)

(kJ mol-1)

(kJ (mol·K)-1)

-11.2

0.0845

∆G° = –RTlnb ∆G° =∆H° - T∆S° 35